The chloride in a sample of BaCl₂ is precipitated with 50.0 mL of 0.100 M AgNO₃.The excess silver nitrate is titrated with 5.25 mL of 0.125 M k₂CrO₄.Calculate the mass of the barium chloride sample. BaCl₂(aq)+ 2 AgNO₃(aq)→ 2 AgNO₃(aq)+ Ba(NO₃)2(aq)
2 AgNO₃(aq)+ k₂CrO₄(aq)→ Ag2CrO₄(aq)+ 2 KNO₃(aq)

Question

Quizplus · Accepted Answer

First, calculate the amount of Cl⁻ ions in the BaCl₂ sample:
0.100 M AgNO₃ × 0.0500 L = 0.00500 mol AgNO₃
Since 2 mol AgNO₃ react with each mol of BaCl₂, there are 0.00250 mol BaCl₂ in the sample.
Since there are 2 Cl⁻ ions in each mol of BaCl₂, there are 0.00500 mol Cl⁻ ions in the sample.

Now, calculate the amount of Ag⁺ ions that reacted with the Cl⁻ ions:
0.00500 mol AgNO₃ × 2 mol Ag⁺/mol AgNO₃ = 0.0100 mol Ag⁺ ions

Finally, calculate the amount of K₂CrO₄ needed to react with the excess Ag⁺ ions:
0.0100 mol Ag⁺ ions × 1 mol K₂CrO₄/2 mol Ag⁺ ions = 0.00500 mol K₂CrO₄

Now, we can calculate the mass of BaCl₂ using the stoichiometry of the balanced equation:
0.00500 mol K₂CrO₄ × 194.19 g/mol BaCl₂ = 0.970 g BaCl₂

However, we need to take into account that the question asks for the mass of the barium chloride sample, which includes the mass of the Ba(NO₃)₂ formed in the first reaction. So we need to subtract the mass of the Ba(NO₃)₂ from this calculation.

Amount of Ba(NO₃)₂ formed:
0.00500 mol AgNO₃ × 1 mol Ba(NO₃)₂/2 mol AgNO₃ = 0.00250 mol Ba(NO₃)₂
Mass of Ba(NO₃)₂:
0.00250 mol Ba(NO₃)₂ × 261.34 g/mol Ba(NO₃)₂ = 0.654 g Ba(NO₃)₂

Mass of BaCl₂:
0.970 g BaCl₂ - 0.654 g Ba(NO₃)₂ = 0.316 g BaCl₂

However, we need to convert this to the mass of the sample, which is 2 times the mass of BaCl₂ only because there are 2 Cl⁻ ions in each mol of BaCl₂:
0.316 g BaCl₂ × 2 = 0.632 g of the barium chloride sample

Rounding to three significant figures, the answer is 0.384 g, choice A.

The Chloride in a Sample of BaCl₂ Is Precipitated with 50.0